Which of the following options is/are true?

Final assessment: The statements "Some k-regular graph can be complete graph" and "In a simple connected graph with more than 2 vertices, it must have at least 2 vertices with same degree." are correct. The statements "A graph G is bipartite if and only has even cycle." and "The degree sum of odd degree vertices must be odd." are incorrect.

• Statement: Some k-regular graph can be complete graph — True. A k-regular graph on n vertices is complete exactly when k = n − 1, because every vertex must be adjacent to all other n − 1 vertices. Example: the complete graph on 4 vertices (K4) is 3-regular.

• Statement: In a simple connected graph with more than 2 vertices, it must have at least 2 vertices with same degree — True. For a connected simple graph with n > 2 vertices, possible degrees are 1,2,...,n−1 (only n−1 distinct values) while there are n vertices. By the pigeonhole principle, two vertices must share the same degree.

• Statement: A graph G is bipartite if and only has even cycle — False. The correct characterization is: a graph is bipartite if and only if it has no odd-length cycles. A bipartite graph need not contain any cycle at all (for example, any tree is bipartite), so requiring an even cycle is incorrect.

• Statement: The degree sum of odd degree vertices must be odd — False. The handshaking lemma implies that the number of vertices with odd degree is even, so the sum of degrees of all odd-degree vertices is even. For example, two vertices of degree 1 have sum 2.